Problem: Simplify and expand the following expression: $ \dfrac{z}{z - 2}-\dfrac{3z + 4}{5z - 4} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(z - 2)(5z - 4)$ Multiply the first term by $\dfrac{5z - 4}{5z - 4}$ $ \begin{align*} \dfrac{z}{z - 2} \times \dfrac{5z - 4}{5z - 4} & = \dfrac{(z)(5z - 4)}{(z - 2)(5z - 4)} \\ & = \dfrac{5z^2 - 4z}{(z - 2)(5z - 4)}\end{align*} $ Multiply the second term by $\dfrac{z - 2}{z - 2}$ $ \begin{align*} \dfrac{3z + 4}{5z - 4} \times \dfrac{z - 2}{z - 2} & = \dfrac{(3z + 4)(z - 2)}{(5z - 4)(z - 2)} \\ & = \dfrac{3z^2 - 2z - 8}{(5z - 4)(z - 2)}\end{align*} $ Now we have: $ = \dfrac{5z^2 - 4z}{(z - 2)(5z - 4)} - \dfrac{3z^2 - 2z - 8}{(5z - 4)(z - 2)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{5z^2 - 4z - (3z^2 - 2z - 8)}{(z - 2)(5z - 4)} $ $ = \dfrac{5z^2 - 4z - 3z^2 + 2z + 8}{(z - 2)(5z - 4)} $ $ = \dfrac{2z^2 - 2z + 8}{(z - 2)(5z - 4)}$ Expand the denominator: $ = \dfrac{2z^2 - 2z + 8}{5z^2 - 14z + 8}$